Prove $\int_{0}^{\pi}\frac{u}{1-\cos u}\ln\frac{1+\sin u}{1-\sin u}{d}u=\left(\pi+2\ln2\right)\pi$ 
How to prove
$$\displaystyle\int_{0}^{\pi}\frac{u}{1-\cos u}\ln\left(\frac{1+\sin u}{1-\sin u}\right)\mathrm{d}u=\left(\pi+2\ln2\right)\pi\,\,?$$

I tried to apply the Feynman method to get the improper integral, but I failed.
Now I am stuck in this improper integral.
This is not a duplicate problem, I am looking for a proof without using complex analysis.
 A: Substitute $u=y+\frac\pi2$ and then split the integral
\begin{align}
I=& \int_{0}^{\pi}\frac{u}{1-\cos u}\ln \frac{1+\sin u}{1-\sin u} \>{d}u
=\int_{0}^{\pi/2}\left( \frac{\frac\pi2+y}{1+\sin y}+ \frac{\frac\pi2-y}{1-\sin y}\right)\ln\frac{1+\cos y}{1-\cos y}\>dt\\
\end{align}
Substitute $t=\tan\frac y2$
\begin{align}
I=& -8\int_{0}^1 \frac{(\frac\pi4 - \tan^{-1} t)\ln t}{(1-t)^2} +\frac{(\frac\pi4 + \tan^{-1} t)\ln t}{(1+t)^2} \> dt
= 4\pi\ln2- 32\int_0^1 \frac{(\frac\pi4 - \tan^{-1} t)\> t \ln t}{(1-t^2)^2}dt\tag1
\end{align}
where the first term results from $ -4\pi\int_0^1 \frac{\ln t}{(1+t)^2}dt=4\pi\ln2$. Integrate the remaining integral by parts
\begin{align}
\int_0^1 \frac{(\frac\pi4 - \tan^{-1} t) t \ln t}{(1-t^2)^2}dt
=&\int_0^1 (\frac\pi4 - \tan^{-1} t)\>d\left( \frac{(1-t^2) \ln(1-t^2) +2t^2\ln t}{4(1-t^2)}\right)\\
 =& \frac14 \int_0^1 \frac{\ln t}{1-t^2}dt +\frac14 \int_0^1
\frac{\ln(1-t^2)-\ln t}{1+t^2}dt
\end{align}
where $\int_0^1 \frac{\ln t}{1-t^2}dt =-\frac{\pi^2}8$
$$K=\int_0^1
\frac{\ln(1-t^2)-\ln t}{1+t^2}dt \overset{t\to\frac{1-t}{1+t}}=
\int_0^1 \frac{2\ln2}{1+t^2}dt -K= \int_0^1 \frac{\ln2}{1+t^2}dt
=\frac\pi4\ln2
$$
Substitute above results into (1) to obtain
$$I= {\pi^2}+2\pi\ln2$$
